Question

有人能告诉我编译时错误的含义并帮助我吗？

其中一个错误：

C：\ doc1 \ collect2.exe [错误] ld返回1退出状态

这是文本文件：

 T(F)       R1        R2       R3        R4
 95.0     95.20     66.10     43.10     29.00
 96.0     96.10     67.60     43.50     31.20
 97.0     97.40     67.00     43.70     30.50
 98.0     97.20     69.10     44.10     30.70
 99.0     98.90     68.00     44.70     32.80
100.0    99.50     71.10     45.10     31.50
101.0   101.00     71.20     45.30     31.60
102.0   101.60     71.00     45.70     30.50
103.0   101.80     73.10     46.30     32.50
104.0   103.70     73.50     46.60     32.70
105.0   105.60     72.80     47.10     33.60

这是我的来源

#include <fstream>
#include <iostream>
#include <string>
#include <iomanip>
#include <cmath>
#include <stdlib.h>

using namespace std;

void read ();
void slope(double,double,double,double,double,int,double);
void write(double,double,double,double,double,double);
double temp, R1, R2, R3, R4,rsq;

int main ()
{
    read();
    return 0;   
}

void read()
{

    ifstream indata("c:\\doc1\\temperaturedata.txt");
    int indx=0;
    if(indata == NULL)
    {
        cout<< "There is no file go get it from the CD"<< endl;
        return;
    }
    const int columns = 5;
    const int rows = 11;

    double Ihatearray[rows][columns];
    string temparature;
    getline(indata, temparature);
    for(int i = 0; i < rows; ++i)
    {
        for (int j = 0; j < columns; ++j)
            {
                cout << "\t" << flush;
                indata >> Ihatearray[i][j];
                cout.setf(ios::fixed);
                cout << setprecision(2) << Ihatearray[i][j] <<flush;
            }
            cout << endl;
            double temp[20], R1[20],R2[20],R3[20],R4[20];
    }
    //indata.close();
    slope(temp, R1, R2, R3, R4,indx,rsq);
}
void slope(double x[5], double y[], double& m, double& b, int n,double& r)
{
    int i;
double xsum = 0., ysum = 0., xave, yave, x2sum = 0., y2sum = 0.,xysum =0.;

for(int i = 0; i < n; i++)
{
    xsum  += x[i];
    ysum  += y[i];
    x2sum += x[i] * x[i];
    y2sum += y[i] * y[i];
    xysum += x[i] * y[i];
}

xave=xsum/n;
yave=ysum/n;

m=(xysum-yave*xsum)/(x2sum-xave*xsum);
b=yave-m*xave;
r=(xysum-n*xave*yave)/(sqrt(x2sum-n*xave*xave)*sqrt(y2sum-n*yave*yave));

write (temp, R1, R2, R3, R4,rsq);

}
void write (double temp [], double R1[], double R2[], double R3[],double R4[],double
            rsq)
{
    //Where I will print out a table of the results.
}

Answer 1

void slope(double,double,double,double,double,int,double);
void write(double,double,double,double,double,double);

和

void slope(double x[5], double y[], double& m, double& b, int n,double& r)

void write (double temp [], double R1[], double R2[], double R3[],double R4[],double rsq)

发现差异？这些声明和定义必须匹配。

更多错误，您的代码看起来像这样

double temp, R1, R2, R3, R4,rsq;

void read()
{
    ...
    for(int i = 0; i < rows; ++i)
    {
        ...
        double temp[20], R1[20],R2[20],R3[20],R4[20];
    }
    //indata.close();
    slope(temp, R1, R2, R3, R4,indx,rsq);
}

现在看看你写的是什么，你已经宣布了temp，R1，R2等两次。您打电话给slope时，您认为哪个声明适用？这是规则，{和}中的声明仅在{和}内可见。

我觉得你的编码应该更有条理。这些只是细节，但细节必须正确。

奇怪的编译错误

1 个答案: